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Difference between unique solution and particular solution

  1. Jan 10, 2013 #1
    1. The problem statement, all variables and given/known data

    For 1st order linear ordinary differential equation, how do "unique solution" and "particular solution" differ?

    2. Relevant equations

    if dy/dx = f(x,y) and partial f(x,y) with respect to y are both continuous, then there exists a unique solution within a region R.



    3. The attempt at a solution

    Not really a homework question. I am just curious and sort of confused. The way I understand it is particular solution is a subset of unique solution, is that true?
     
  2. jcsd
  3. Jan 10, 2013 #2

    SteamKing

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    If a given differential equation f(x,y,y',y'',...) = 0, this is called a homogeneous equation and it has what is called a complementary solution.

    If the differential equation f(x,y,y',y'',...) = c + f(x,y), this is called a non-homogeneous equation. The solution to these types of equations are known as particular solutions.

    If the ODE is linear, then the solution to the equation will be a linear combination of the complementary and the particular solutions.

    If there is one solution for a given ODE with given boundary conditions, then that solution is called unique.
     
  4. Jan 10, 2013 #3

    HallsofIvy

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    The differ only in emphasis. If the general solution to a differential equation is, say, [itex]y= Ae^x+ Be^{-x}[/itex], then "a specific solution" is any solution with a specific choice of A and B. The "unique solution" is a specific solution that satisfies given initial conditions.

    For example, if the differential equation is [itex]d^2y/dx^2= y[/itex] then the "general solution' is, as before, [itex]y= Ae^x+ Be^{-x}[/itex] for any constants A and B. A specific solution (notice the use of the indefinite article, "a") might be [itex]y= e^x- 2e^{-x}[/itex] where I have arbitrarily chosen A= 1, B= -2. The unique solution (notice the use of the definite article, "the") to the differential equation, [itex]d^2y/dx^2= y[/itex] with the initial conditions y(0)= 1, y'(0)= 0, is [itex](1/2)e^x+ (1/2)e^{-x}[/itex].

    Of course, the "unique solution" to the given "initial value problem" is one of the infinite number of "specific solutions" to the differential equation.
     
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